w23-logic-3/inputs/lecture_15.tex

\lecture{15}{2023-12-05}{}

\begin{theorem}[The Boundedness Theorem]
    \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}

    Let $X$ be Polish, $C \subseteq X$ coanalytic,
    $\phi\colon  C \to \omega_1$ a coanalytic rank on $C$,
    $A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$.
    Then $\sup \{\phi(x) : x \in A\} < \omega_1$.

    Moreover for all $\xi < \omega_1$,
    \[
    D_\xi \coloneqq  \{x \in C : \phi(x) < \xi\}
    \]
    and
    \[
    E_\xi \coloneqq  \{x \in C : \phi(x) \le \xi\} 
    \]
    are Borel subsets of $X$.
\end{theorem}
\begin{proof}
%     Let
%     \begin{IEEEeqnarray*}{rCl}
%         x \prec y&:\iff& x,y \in A \land \phi(x) < \phi(y)\\
%                  &\iff& x,y \in A \land y \not\le_\phi^\ast x.
%     \end{IEEEeqnarray*}
%     Since $A$ is analytic,
%     this relation is analytic and wellfounded on $X$.
%     By \yaref{thm:kunenmartin}
%     we get $\rho(\prec) < \omega_1$.
%     Thus $\sup \{\phi(x) : x \in A\} < \omega_1$.
% 
%     Since $D_\xi = \bigcup_{\eta < \xi} E_\xi$,
%     it suffices to check $E_\xi \in \Sigma_1^1(X)$.
%     If $\alpha = \sup \{\phi(x) : x \in C\}$,
%     we have $E_\xi = E_\alpha$ for all $\alpha < \xi < \omega_1$.

    \todo{TODO: Copy from official notes}
\end{proof}

\pagebreak
\section{Abstract Topological Dynamics}
Recall:
\begin{definition}+
    Let $X$ be a set.
    A \vocab{group action} of a group $G$ on $X$
    is a function
    $\alpha\colon G \times X \to X$
    such that
    \begin{itemize}
        \item $\forall x \in X.~\alpha(1_G,x) = x$,
        \item $\forall g,h \in G, x \in X.~\alpha(gh,x) = \alpha(g,\alpha(h,x))$.
    \end{itemize}

    Often we will abbreviate $\alpha(g,x)$ as $g\cdot x$.
\end{definition}
\begin{remark}+
    Group actions of a group $G$ on a set $X$
    correspond to group-homomorphisms
    $G \to \Sym(X)$.
    Indeed for a group action $\alpha\colon G \times X \to X$
    consider
    \begin{IEEEeqnarray*}{rCl}
        G&\longrightarrow & \Sym(X) \\
        g&\longmapsto & (x \mapsto g \cdot x).
    \end{IEEEeqnarray*}
\end{remark}



\begin{definition}
    Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology}
    and let $X$ be a compact metrizable space.

    A \vocab{flow}  $(X, T)$, sometimes denoted $T \acts X$
    is a continuous action
    \begin{IEEEeqnarray*}{rCl}
        T \times X&\longrightarrow & X \\
        (t,x) &\longmapsto & tx.
    \end{IEEEeqnarray*}

    A flow is \vocab{minimal}  iff every orbit is dense.

    $(Y,T)$ is a \vocab{subflow}  of $(X,T)$ if $Y \subseteq X$
    and $Y$ is invariant under $T$,
    i.e.~$\forall t \in T,y \in Y.~ty \in Y$.

    A flow $(X,T)$ is \vocab{isometric}
    iff there is a metric $d$ on $X$ such that
    for all $t \in T$ the map
    \begin{IEEEeqnarray*}{rCl}
        a_t\colon X &\longrightarrow & X \\
        x &\longmapsto & tx
    \end{IEEEeqnarray*}
    is an \vocab{isometry},
    i.e.~$\forall t \in T.~\forall x,y \in X.~d(a_t(x),a_t(y)) = d(x,y)$.

    If $(X,T)$ is a flow, then a pair $(x,y)$, $x \neq y$
    is \vocab{proximal} iff
    \[
    \exists  z \in X.~\exists (t_n)_{n < } \in T^{\omega}.~t_n x \xrightarrow{n \to \infty}  z \land t_n y \xrightarrow{n \to \infty} z.
    \]

    A flow is \vocab{distal} iff
    it has no proximal pair.
\end{definition}
\begin{remark}
    Note that a flow is minimal iff it has no proper subflows.
\end{remark}
% \begin{example}
%     Recall that $S_1 = \{z \in \C : |z| = 1\}$.
%     Let $X = S_1$, $T = S_1$
%     $(\alpha,\beta) \mapsto \alpha + \beta$ is isometric.
%     % TODO: In the official notes it says \alpha + \beta, but this is no group action.
%     % Maybe \alpha * \beta ?
% \end{example}

\begin{definition}
    Let $X,Y$ be compact metric spaces
    and $\pi\colon (X,T) \to (Y,T)$ a factor map.
    Then $(X,T)$ is an \vocab{isometric extension} 
    of $(Y,T)$ if there is a real valued $\rho$
    defined on $\{(x_1,x_2) \in X^2 : \pi(x_1) = \pi(x_2)\}$
    such that
    \begin{enumerate}[(a)]
        \item $\rho$ is continuous.
        \item For each $y \in Y$, $\rho$ is a metric on the fibre
            $X_y \coloneqq \{x \in X: \pi(x) = y\}$.
        \item $\forall t \in T.~\rho(tx_1,tx_2) = \rho(x_1,x_2)$.
        \item $\forall y,y' \in Y.~$
            the metric spaces $(X_y, \rho)$ and  $(X_{y'}, \rho)$
            are isometric.
    \end{enumerate}
\end{definition}
\begin{remark}
    A flow is isometric iff it is an isometric extension
    of the trivial flow,
    i.e.~the flow acting on a singleton.
    % TODO THINK ABOUT THIS!
\end{remark}
\begin{proposition}
    An isometric extension of a distal flow is distal.
\end{proposition}
\begin{proof}
    Let $\pi\colon  X\to Y$ be an isometric extension.
    Towards a contradiction,
    suppose that $x_1,x_2 \in X$ are proximal.
    Take $z \in X$ and $(g_n) \in T^{\omega}$
    such that $g_n x_1 \to z$ and $g_n x_2 \to z$.

    Then $g_n \pi(x_1) \to  \pi(z)$
    and $g_n \pi(x_2) \to \pi(z)$,
    so by distality of $Y$
    we have $\pi(x_1) = \pi(x_{2})$.
    Then $\rho(g_n x_1, g_n x_2)$
    is defined and equal to $\rho(x_1,x_2)$.
    By the continuity of $\rho$,
    we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$.
    Therefore $\rho(x_1,x_2) = 0$.
    Hence $x_1 = x_2$ $\lightning$.
\end{proof}

\begin{definition}
    Let $\Sigma = \{(X_i, T) : i \in I\} $
    be a collection of factors of $(X,T)$. % TODO State precise definition of a factor
    Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
    Then $(X, T)$ is the \vocab{limit}  of $\Sigma$
    iff
    \[
    \forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
    \]
    % TODO think about abstract nonsense
\end{definition}

\begin{proposition}
    A limit of distal flows is distal.
\end{proposition}
\begin{proof}
    Let $(X,T)$ be a limit of $\Sigma = \{(X_i, T) : i \in I\}$.
    Suppose that each $(X_i, T)$ is distal.
    If $(X,T)$ was not distal,
    then there were $x_1, x_2, z \in X$
    and a sequence $(g_n)$ in $T$
    with $g_n x_1 \to z$ and $g_n x_2 \to z$.
    Take $i \in I$ such that $\pi_i(x_1) \neq \pi_i(x_2)$.
    But then $g_n \pi_i(x_1) \to \pi_i(z)$
    and $g_n \pi_i(x_2) \to \pi_i(z)$,
    which is a contradiction since $(X_i, T)$ is distal.
\end{proof}